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Statistics - Measures of Dispersion: Range, Mean Deviation

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Dispersion refers to the extent to which numerical data is likely to vary about an average value. Visually, if you plot data points on a horizontal number line, a low dispersion means the points are tightly clustered together, while a high dispersion means they are spread far apart across the line.

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The Range is the simplest measure of dispersion, calculated as the difference between the maximum and minimum observations in a data set. Graphically, the range represents the total length of the interval on the x-axis covered by the data points.

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Mean Deviation (M.D.) measures the average of the absolute differences between each data point and a central value (usually the mean or median). We use absolute values (modulus) because the sum of actual deviations from the mean is always zero; visually, taking the absolute value ensures we are measuring the total distance from the center regardless of direction.

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Mean Deviation about Mean for Ungrouped Data is found by calculating the arithmetic mean xˉ\bar{x}, then finding the average of the absolute differences ∣xi−xˉ∣|x_i - \bar{x}|. This tells us, on average, how far each observation lies from the center of the distribution.

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Mean Deviation about Median is often preferred when the data contains extreme outliers, as the median is less affected by values at the far ends of the distribution. In a visual frequency curve, the median represents the point that divides the area under the curve into two equal halves, and M.D. about the median measures the average 'distance' from this dividing line.

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For Discrete Frequency Distributions, the formula incorporates frequencies fif_i as weights. Each absolute deviation ∣xi−a∣|x_i - a| is multiplied by its corresponding frequency, reflecting how many times that specific distance from the center occurs in the dataset.

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For Continuous Frequency Distributions, we use the class marks (mid-points of class intervals) as the xix_i values. Visually, this assumes that the data within each rectangular bar of a histogram is centered at the midpoint of that bar's width.

📐Formulae

Range = Xmax−XminX_{max} - X_{min}

Mean for ungrouped data: xˉ=∑xin\bar{x} = \frac{\sum x_i}{n}

Mean Deviation about Mean (Ungrouped): M.D.(xˉ)=∑∣xi−xˉ∣nM.D.(\bar{x}) = \frac{\sum |x_i - \bar{x}|}{n}

Mean Deviation about Median (Ungrouped): M.D.(M)=∑∣xi−M∣nM.D.(M) = \frac{\sum |x_i - M|}{n}

Mean Deviation about Mean (Grouped): M.D.(xˉ)=∑fi∣xi−xˉ∣NM.D.(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{N}, where N=∑fiN = \sum f_i

Mean Deviation about Median (Grouped): M.D.(M)=∑fi∣xi−M∣NM.D.(M) = \frac{\sum f_i |x_i - M|}{N}, where N=∑fiN = \sum f_i

Median for Continuous Distribution: M=l+(N2−Cf)×hM = l + \left( \frac{\frac{N}{2} - C}{f} \right) \times h

💡Examples

Problem 1:

Find the Mean Deviation about the mean for the following data: 6,7,10,12,13,4,8,126, 7, 10, 12, 13, 4, 8, 12.

Solution:

  1. Calculate the Mean (xˉ)(\bar{x}): xˉ=6+7+10+12+13+4+8+128=728=9\bar{x} = \frac{6+7+10+12+13+4+8+12}{8} = \frac{72}{8} = 9
  2. Calculate absolute deviations ∣xi−xˉ∣|x_i - \bar{x}|: ∣6−9∣=3,∣7−9∣=2,∣10−9∣=1,∣12−9∣=3,∣13−9∣=4,∣4−9∣=5,∣8−9∣=1,∣12−9∣=3|6-9|=3, |7-9|=2, |10-9|=1, |12-9|=3, |13-9|=4, |4-9|=5, |8-9|=1, |12-9|=3
  3. Sum of absolute deviations: ∑∣xi−xˉ∣=3+2+1+3+4+5+1+3=22\sum |x_i - \bar{x}| = 3+2+1+3+4+5+1+3 = 22
  4. Calculate Mean Deviation: M.D.(xˉ)=228=2.75M.D.(\bar{x}) = \frac{22}{8} = 2.75

Explanation:

To find the mean deviation, we first establish the average (mean) as the central point. Then, we determine how far each individual number is from that 9. By averaging these distances, we conclude that on average, the data points deviate by 2.75 units from the mean.

Problem 2:

Calculate the Mean Deviation about the median for the data: 3,9,5,3,12,10,18,4,7,19,213, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21.

Solution:

  1. Arrange data in ascending order: 3,3,4,5,7,9,10,12,18,19,213, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21
  2. Find the Median (MM): Since n=11n=11 (odd), M=(11+12)thM = (\frac{11+1}{2})^{th} observation = 6th6^{th} observation = 99.
  3. Calculate absolute deviations ∣xi−9∣|x_i - 9|: ∣3−9∣=6,∣3−9∣=6,∣4−9∣=5,∣5−9∣=4,∣7−9∣=2,∣9−9∣=0,∣10−9∣=1,∣12−9∣=3,∣18−9∣=9,∣19−9∣=10,∣21−9∣=12|3-9|=6, |3-9|=6, |4-9|=5, |5-9|=4, |7-9|=2, |9-9|=0, |10-9|=1, |12-9|=3, |18-9|=9, |19-9|=10, |21-9|=12
  4. Sum of deviations: ∑∣xi−M∣=6+6+5+4+2+0+1+3+9+10+12=58\sum |x_i - M| = 6+6+5+4+2+0+1+3+9+10+12 = 58
  5. M.D.(M)=5811≈5.27M.D.(M) = \frac{58}{11} \approx 5.27

Explanation:

When calculating M.D. about the median, the first step is always sorting the data to find the middle value. The median here is 9. We then sum the absolute distances of all points from 9 and divide by the count of observations (11).